1
JEE Main 2022 (Online) 29th June Morning Shift
+4
-1

Let $$f:R \to R$$ be a function defined by :

$$f(x) = \left\{ {\matrix{ {\max \,\{ {t^3} - 3t\} \,t \le x} & ; & {x \le 2} \cr {{x^2} + 2x - 6} & ; & {2 < x < 3} \cr {[x - 3] + 9} & ; & {3 \le x \le 5} \cr {2x + 1} & ; & {x > 5} \cr } } \right.$$

where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and $$I = \int\limits_{ - 2}^2 {f(x)\,dx}$$. Then the ordered pair (m, I) is equal to :

A
$$\left( {3,\,{{27} \over 4}} \right)$$
B
$$\left( {3,\,{{23} \over 4}} \right)$$
C
$$\left( {4,\,{{27} \over 4}} \right)$$
D
$$\left( {4,\,{{23} \over 4}} \right)$$
2
JEE Main 2022 (Online) 29th June Morning Shift
+4
-1

The area enclosed by y2 = 8x and y = $$\sqrt2$$ x that lies outside the triangle formed by y = $$\sqrt2$$ x, x = 1, y = 2$$\sqrt2$$, is equal to:

A
$${{16\sqrt 2 } \over 6}$$
B
$${{11\sqrt 2 } \over 6}$$
C
$${{13\sqrt 2 } \over 6}$$
D
$${{5\sqrt 2 } \over 6}$$
3
JEE Main 2022 (Online) 29th June Morning Shift
+4
-1

$$\int_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx}$$,

where [t] denotes greatest integer less than or equal to t, is equal to:

A
$$-$$3
B
$$-$$2
C
2
D
0
4
JEE Main 2022 (Online) 28th June Evening Shift
+4
-1

Let f : R $$\to$$ R be a differentiable function such that $$f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0$$ and $$f'\left( {{\pi \over 2}} \right) = 1$$ and

let $$g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt}$$ for $$x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)$$. Then $$\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)$$ is equal to :

A
2
B
3
C
4
D
$$-$$3
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